Question

A right-angled prism (${45}^{\circ },{90}^{\circ },{45}^{\circ }$) of refractive index n has a plate of refractive index $n_1(n_1<n)$cemented to its diagonal face. The assembly is in the air. A ray is an incident on AB. If the angle of incidence at AB for which the ray strikes the diagonal face at the critical angle X in degrees. Find X.

Formula used:
Snell’s law tells us:
$$n_i\sin {\theta }_i=n_r\sin {\theta }_r$$ ………... (1)
Where,
$$n_i$$ is the refractive index of the medium light is coming from,
$${\theta }_i$$ is the angle of incidence,
$$n_r$$ is the refractive index of the medium light is going into,
$${\theta }_r$$ is the angle of refraction.

Answer 1

Hint: When light falls on the surface AB, after refraction it passes into the prism. Due to refraction the light beam bends towards the normal as the refractive index of the prism is greater than that of the air. Then this ray falls on the other surface at a critical angle; it gazes through the diagonal surface.

Complete Step by Step Answer:
Given:
The prism is a right angled prism, i.e. $\mathrm{\angle }A={45}^{\circ }$.
Refractive index of prism is n and that of the plate is $n_1$.
Angle of incidence of the ray at AB is $X$.

To find: The value of $X$.
Step 1
For refraction at the surface AB we have, $${\theta }_i=X,{\theta }_r=r,n_i=1,n_r=n$$. So, use these values in eq.(1) to get:
$$\begin{gathered} 1\times \sin X=n\times \sin r \\ \therefore \sin X=n\sin r \end{gathered}$$ ………………….(2)
Step 2
In the AC surface we have, $${\theta }_i={\theta }_C,{\theta }_r={90}^{\circ },n_i=n,n_r=n_1$$ where, $${\theta }_C$$ is the critical angle for AC surface. Now, use these values in eq.(1) to get:
$$\begin{gathered} n\times \sin {\theta }_C=n_1\times \sin {90}^{\circ } \\ \Rightarrow \sin {\theta }_C={\displaystyle \frac{n_1}{n}} \\ \Rightarrow {\theta }_C={\sin }^{-1}\left({\displaystyle \frac{n_1}{n}}\right) \end{gathered}$$ .....................(3)
Step 3
Now, focus on the quadrilateral AOPQ. For this quadrilateral we have $\mathrm{\angle }AOP=\mathrm{\angle }AQP={90}^{\circ }$and $\mathrm{\angle }A={45}^{\circ }$. Hence, we get $$\mathrm{\angle }OPQ={360}^{\circ }-({90}^{\circ }+{90}^{\circ }+{45}^{\circ })={135}^{\circ }$$.
For the triangle OPQ we have $$\mathrm{\angle }OPQ={135}^{\circ }$$. Hence, we get:
$$\begin{gathered} \mathrm{\angle }POQ+\mathrm{\angle }PQO={180}^{\circ }-\mathrm{\angle }OPQ \\ \Rightarrow r+{\theta }_C={180}^{\circ }-{135}^{\circ } \\ \therefore r={45}^{\circ }-{\theta }_C \end{gathered}$$.................4)
Step 4:
Substitute the values of r from eq.(4) and that of $${\theta }_C$$ from eq.(3) into eq.(2) to get the final relationship as:
$$\begin{gathered} \sin X=n\sin \left({45}^{\circ }-{\theta }_C\right) \\ \therefore X={\sin }^{-1}\left[n\sin \left({45}^{\circ }-{\sin }^{-1}\left({\displaystyle \frac{n_1}{n}}\right)\right)\right] \end{gathered}$$

$\therefore$ The value of X is $${\sin }^{-1}\left[n\sin \left({45}^{\circ }-{\sin }^{-1}\left({\displaystyle \frac{n_1}{n}}\right)\right)\right]$$.

Note:
If you carefully notice the obtained incident angle term you can notice that here, $$n_1$$ and n should be chosen in such a way that $${\sin }^{-1}\left({\displaystyle \frac{n_1}{n}}\right)$$ term be less than $${45}^{\circ }$$. Otherwise, the top angle $\mathrm{\angle }A$ can be further increased to get a proper incident angle X for which the ray will fall at a critical angle on AC surface.